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Is there a function f(a,b) which maps ordered pairs to lines in a plane in a continuous, bijective manner?

Here is the definition I am using for the limit with lines: a sequence of lines $L(1), L(2), \dots$ is said to approach another line $L$ if, for any point $p$ on $L$, the limit as $n\to\infty$ of the distance between $p$ and $L(n)$ is 0.

If there is no such function, can anyone think of a proof?

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Which topology do you put on the target set? –  Qfwfq Jan 2 '11 at 17:01
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(a,b) |-> {y = ax+b} ? –  Allen Knutson Jan 2 '11 at 17:04
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MO is intended for research-level questions. The FAQ lists a number of websites that are more appropriate for elementary questions (for instance, math.stackexchange.com). –  Andy Putman Jan 2 '11 at 17:08
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Well I guess yrudoy doesn't specifically ask for his function to be surjective... –  Steven Gubkin Jan 2 '11 at 17:12
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1 Answer 1

up vote 10 down vote accepted

There is no such bijection.

A line in the plane is almost the same as a plane through the origin in 3-space (by intersecting with the plane at height 1), except there's one plane through the origin that doesn't give you a line (the z=0 plane). So the space of lines in the plane is homeomorphic to $\mathbb{RP}^2$ minus a point: an open mobius strip! So the question is asking if there is a continuous bijection from the open disk $D$ to the open mobius strip $M$. Invariance of domain implies that a continuous bijection between manifolds of the same dimension is a homeomorphism. $D$ and $M$ are not homeomorphic, so there cannot be a continuous bijection between them.

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(copied from meta) Very nice, but I believe a continuous map needs to be proper for the induced function on one-point compactifications is also continuous. So you have shown that there is no proper map of the desired type, but I think maybe a bit more is required to show there is no continuous map at all? –  Noah Stein Jan 3 '11 at 1:09
    
You're right. I don't see a fix right now. –  Anton Geraschenko Jan 3 '11 at 1:14
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Invariance of domain says that a continuous, one-to-one map of manifolds (without boundary) is a homeomorphism onto its image, so I think that finishes the proof because $D$ (contractible) is not homeomorphic to the open mobius strip (not contractible). –  Noah Stein Jan 3 '11 at 1:22
    
(there should have been an "of the same dimension" after the word "manifolds") –  Noah Stein Jan 3 '11 at 1:22
    
Very nice. I'll edit the answer now. I was sitting around trying to prove invariance of domain. Is there an easy way to see that a continuous bijection $\mathbb R^n\to \mathbb R^n$ is a homeomorphism? –  Anton Geraschenko Jan 3 '11 at 1:30

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